MHT CET · Physics · Center of Mass Momentum and Collision
Using variation of force and time given below, final velocity of a particle of mass \(2 \mathrm{~kg}\) moving with initial velocity \(6 \mathrm{~m} / \mathrm{s}\) will be
- A \(10 \mathrm{~m} / \mathrm{s}\)
- B \(5 \mathrm{~m} / \mathrm{s}\)
- C \(12 \mathrm{~m} / \mathrm{s}\)
- D \(0 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(C) \(12 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Now, using the impulse-momentum theorem to find the final velocity
\(\begin{aligned}& 12=2 \times(-6) \\& v-6=\frac{12}{2}=6 \\& v=12 \mathrm{~m} / \mathrm{s}\end{aligned}\)
So, with the given force-time data \((0,6)\) and \((4,0)\), the final velocity of the particle is \(12 \mathrm{~m} / \mathrm{s}\)
\(\begin{aligned}& 12=2 \times(-6) \\& v-6=\frac{12}{2}=6 \\& v=12 \mathrm{~m} / \mathrm{s}\end{aligned}\)
So, with the given force-time data \((0,6)\) and \((4,0)\), the final velocity of the particle is \(12 \mathrm{~m} / \mathrm{s}\)
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